Problem: Graph this system of equations and solve. $2x+10y = 40$ $-8x+10y = -10$ $1$ $2$ $3$ $4$ $5$ $6$ $7$ $8$ $9$ $10$ $\llap{-}2$ $\llap{-}3$ $\llap{-}4$ $\llap{-}5$ $\llap{-}6$ $\llap{-}7$ $\llap{-}8$ $\llap{-}9$ $\llap{-}10$ $1$ $2$ $3$ $4$ $5$ $6$ $7$ $8$ $9$ $10$ $\llap{-}2$ $\llap{-}3$ $\llap{-}4$ $\llap{-}5$ $\llap{-}6$ $\llap{-}7$ $\llap{-}8$ $\llap{-}9$ $\llap{-}10$ Click and drag the points to move the lines.
Solution: Convert the first equation, $2x+10y = 40$ , to slope-intercept form. $y = -\dfrac{1}{5} x + 4$ The y-intercept for the first equation is $4$ , so the first line must pass through the point $(0, 4)$ The slope for the first equation is $-\dfrac{1}{5}$ . Remember that the slope tells you rise over run. So in this case for every $1$ position you move down (because it's negative) You must also move $5$ positions to the right. $5$ positions to the right. Graph the blue line so it passes through $(0, 4)$ and $(5, 3)$ Convert the second equation, $-8x+10y = -10$ , to slope-intercept form. $y = \dfrac{4}{5} x - 1$ The y-intercept for the second equation is $-1$ , so the second line must pass through the point $(0, -1)$ The slope for the second equation is $\dfrac{4}{5}$ . Remember that the slope tells you rise over run. So in this case for every $4$ positions you move up You must also move $5$ positions to the right. $5$ positions to the right. $4$ positions up from $(0, -1)$ is $(5, 3)$ Graph the green line so it passes through $(0, -1)$ and $(5, 3)$ The solution is the point where the two lines intersect. The lines intersect at $(5, 3)$.